Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 2x}{x + 6} = \dfrac{6x + 60}{x + 6}$
Answer: Multiply both sides by $x + 6$ $ \dfrac{x^2 + 2x}{x + 6} (x + 6) = \dfrac{6x + 60}{x + 6} (x + 6)$ $ x^2 + 2x = 6x + 60$ Subtract $6x + 60$ from both sides: $ x^2 + 2x - (6x + 60) = 6x + 60 - (6x + 60)$ $ x^2 + 2x - 6x - 60 = 0$ $ x^2 - 4x - 60 = 0$ Factor the expression: $ (x + 6)(x - 10) = 0$ Therefore $x = -6$ or $x = 10$ At $x = -6$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -6$, it is an extraneous solution.